You have found the following ages (in years) of all 5 tigers at your local zoo: $ 16,\enspace 1,\enspace 13,\enspace 7,\enspace 12$ What is the average age of the tigers at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 5 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{16 + 1 + 13 + 7 + 12}{{5}} = {9.8\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $16$ years $6.2$ years $38.44$ years $^2$ $1$ year $-8.8$ years $77.44$ years $^2$ $13$ years $3.2$ years $10.24$ years $^2$ $7$ years $-2.8$ years $7.84$ years $^2$ $12$ years $2.2$ years $4.84$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{38.44} + {77.44} + {10.24} + {7.84} + {4.84}} {{5}} $ $ {\sigma^2} = \dfrac{{138.8}}{{5}} = {27.76\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{27.76\text{ years}^2}} = {5.3\text{ years}} $ The average tiger at the zoo is 9.8 years old. There is a standard deviation of 5.3 years.